poj1787 Charlie's Change

标签：背包dp

Description

Input

Output

Sample Input

12 5 3 1 2
16 0 0 0 1
0 0 0 0 0

Sample Output

Throw in 2 cents, 2 nickels, 0 dimes, and 0 quarters.
Charlie cannot buy coffee.
这题可以用多重背包，用dp[m][5]分别记录m价钱刚好能买到的物品个数的最大值，买1元，5元，10元，25元物品的个数。
#include<stdio.h>
#include<string.h>
#define inf 88888888
int max(int a,int b){
	return a>b?a:b;
}
int w[10]={0,1,5,10,25},v[10]={1,1,1,1,1,1};
int dp[10006][6];
int main()
{
	int n,m,i,j,a,b,c,d,k,ans,sum;
	int num[19];
	while(scanf("%d%d%d%d%d",&m,&num[1],&num[2],&num[3],&num[4])!=EOF)
	{
		if(m+num[1]+num[2]+num[3]+num[4]==0)break;
		if(m>num[1]+num[2]*5+num[3]*10+num[4]*25){
			printf("Charlie cannot buy coffee.\n");continue;
		}
		if(num[1]>=m){
			printf("Throw in %d cents, 0 nickels, 0 dimes, and 0 quarters.\n",m);continue;
		}
		for(i=0;i<=m;i++){
			dp[i][0]=-inf;
		}
		dp[0][0]=dp[0][1]=dp[0][2]=dp[0][3]=dp[0][4]=0;
		
		for(i=1;i<=4;i++){
			ans=num[i]*w[i];
			if(ans>=m){
				for(j=w[i];j<=m;j++){
					if(dp[j-w[i]][0]>=0){
						if(dp[j-w[i]][0]+v[i]>dp[j][0]){
							dp[j][0]=dp[j-w[i]][0]+v[i];
							dp[j][1]=dp[j-w[i]][1];dp[j][2]=dp[j-w[i]][2];dp[j][3]=dp[j-w[i]][3];dp[j][4]=dp[j-w[i]][4];
							dp[j][i]++;
							
						}
					}
				}
			}
			else{
				k=1;sum=0;
				while(sum+k<num[i]){
					sum+=k;
					for(j=m;j>=k*w[i];j--){
						if(dp[j-k*w[i]][0]>=0){
							if(dp[j-k*w[i]][0]+k*v[i]>dp[j][0]){
								dp[j][0]=dp[j-k*w[i]][0]+k*v[i];
							    dp[j][1]=dp[j-k*w[i]][1];dp[j][2]=dp[j-k*w[i]][2];dp[j][3]=dp[j-k*w[i]][3];dp[j][4]=dp[j-k*w[i]][4];
							    dp[j][i]+=k;
							    
							}
						}
						
					}
					k=k*2;
				}
				k=num[i]-sum;
				for(j=m;j>=k*w[i];j--){
						if(dp[j-k*w[i]][0]>=0){
							if(dp[j-k*w[i]][0]+k*v[i]>dp[j][0]){
								dp[j][0]=dp[j-k*w[i]][0]+k*v[i];
							    dp[j][1]=dp[j-k*w[i]][1];dp[j][2]=dp[j-k*w[i]][2];dp[j][3]=dp[j-k*w[i]][3];dp[j][4]=dp[j-k*w[i]][4];
							    dp[j][i]+=k;
							  
							}
						}
						
				}
			}
		}
		if(dp[m][0]>0)
		printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",dp[m][1],dp[m][2],dp[m][3],dp[m][4]);
		else printf("Charlie cannot buy coffee.\n");
		//printf("%d\n",dp[m][0]);
	}
	return 0;
}

poj1787 Charlie's Change

标签：背包dp

原文地址：http://blog.csdn.net/kirito_acmer/article/details/46695591